A charged line extends from y = 2.50 cm to y = -2.50 cm. The total charge distributed uniformly along the line is -9.00 nC. Find the magnitude of the electric field on the x-axis at x = 0.25 cm.
The charge per unit length is
-9*10^-9C/0.05m = =1.8*10^-7 C/m
You will have to perform an integration of Coulombs law for the E-field along the x axis at x = 0.25. There will be no y component, due to symmetry.
dE = k*(charge density)*x*dy/(x^2+y^2)^(3/2)
Integrate from y=-0.025 to +.025 m, with x fixed.
To find the magnitude of the electric field on the x-axis at x = 0.25 cm, we can use Coulomb's law which states that the electric field due to a uniformly charged line at a point on the x-axis is given by:
E = (k * λ) / r
where:
E is the electric field,
k is the Coulomb's constant (8.99 x 10^9 N·m^2/C^2),
λ is the linear charge density (charge per unit length) of the line,
and r is the distance of the observation point from the line.
First, let's calculate the linear charge density λ. We know that the total charge Q distributed uniformly along the line is -9.00 nC. The length of the line is 2.50 cm - (-2.50 cm) = 5.00 cm.
So, λ = Q / L
λ = (-9.00 nC) / (5.00 cm)
λ = (-9.00 x 10^-9 C) / (5.00 x 10^-2 m)
λ = -1.80 x 10^-7 C/m
Next, let's calculate the distance r from the line to the point on the x-axis, which is 0.25 cm = 0.0025 m.
Now we can calculate the electric field E using Coulomb's law:
E = (k * λ) / r
E = (8.99 x 10^9 N·m^2/C^2) * (-1.80 x 10^-7 C/m) / (0.0025 m)
Calculating this expression, we get:
E ≈ -129360 N/C
Therefore, the magnitude of the electric field on the x-axis at x = 0.25 cm is approximately 1.29 x 10^5 N/C.
To find the magnitude of the electric field on the x-axis at x = 0.25 cm, we can use Coulomb's law. Coulomb's law states that the electric field created by a charged line is directly proportional to the charge and inversely proportional to the distance.
First, let's find the electric field created by a small section of the line. We'll consider a small section of length dl at a distance y from the origin.
The charge dq associated with this small section is given by:
dq = (total charge / total length) * dl
Given that the total charge distributed uniformly along the line is -9.00 nC and the total length is 5 cm (2.5 cm to -2.5 cm), we can calculate the charge density:
λ = (total charge / total length)
λ = (-9.00 nC / 5 cm)
To find the electric field created by this small section, we use Coulomb's law:
dE = (k * dq) / r^2
Where k is the Coulomb's constant (k ≈ 9.0 x 10^9 Nm^2/C^2) and r is the distance from the small section to the point on the x-axis at x = 0.25 cm.
Considering that r^2 = y^2 + x^2, where y is the distance of the section from the y-axis, we can simplify this expression as:
r = sqrt(y^2 + x^2)
Plugging in the values, we get:
r = sqrt((2.50 cm)^2 + (0.25 cm)^2)
Once we have calculated the value of r, we can calculate the electric field at x = 0.25 cm for a small section by using Coulomb's law:
dE = (k * dq) / r^2
Now, the electric field on the x-axis at x = 0.25 cm is the sum of all the electric fields created by every small section along the charged line. Therefore, we integrate the above equation to find the total electric field. Since the charge is distributed uniformly, we can replace the charge density λ with dq / dl, where dl is an infinitesimally small section of the line.
|dE| = (k * (dq / dl)) / r^2
Integrating both sides of the equation from y = -2.50 cm to y = 2.50 cm:
∫|dE| = ∫ [(k * (dq / dl)) / r^2] from y = -2.50 cm to y = 2.50 cm
Simplifying the equation and substituting the values, we can find the magnitude of the electric field on the x-axis at x = 0.25 cm.